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NIMCET 2021 — Mathematics PYQ

NIMCET | Mathematics | 2021

The eccentric angle of the extremities of latusrectum of the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ are given by

Choose the correct answer:

Explanation

1. Coordinates of the Latus Rectum Extremities

The latus rectum of an ellipse is a chord passing through the focus $(ae, 0)$ and perpendicular to the major axis. The coordinates of the endpoints (extremities) in the first quadrant and fourth quadrant are:

$(x, y) = (ae, \pm \frac{b^2}{a})$

2. Parametric Coordinates and Eccentric Angle

Any point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ can be represented in parametric form as:

$x = a \cos \phi$

$y = b \sin \phi$

Where $\phi$ is the eccentric angle.

3. Equating the Coordinates

To find the eccentric angle $\phi$ for the latus rectum, we equate the parametric forms to the coordinates found in Step 1:

$a \cos \phi = ae \implies \cos \phi = e$

$b \sin \phi = \pm \frac{b^2}{a} \implies \sin \phi = \pm \frac{b}{a}$

4. Finding the Tangent of the Angle

To find $\phi$ in terms of $\tan^{-1}$ , we divide $\sin \phi$ by $\cos \phi$ :

$\tan \phi = \frac{\sin \phi}{\cos \phi} = \frac{\pm b/a}{e} = \pm \frac{b}{ae}$

Therefore:

$\phi = \tan^{-1}\left(\pm \frac{b}{ae}\right)$

5. Simplifying using Eccentricity ( $e$ )

We know that for an ellipse, $b^2 = a^2(1 - e^2)$ , which means $\frac{b}{a} = \sqrt{1 - e^2}$ .

Substituting this into our $\tan \phi$ expression:

$\tan \phi = \frac{\pm \sqrt{1 - e^2}}{e}$

$\phi = \tan^{-1}\left(\frac{\pm \sqrt{1 - e^2}}{e}\right)$

Final Answer

Comparing our result with the given options, we find that the expression matches Option (D).

Correct Option: (D) $\tan^{-1}\left(\frac{\pm \sqrt{1-e^2}}{e}\right)$

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